Simultaneous equations with x and y exponents

Stuck Man · #1 $Old$ November 3rd, 2009, 06:10 AM

Is there a way of doing these?

Jameson · #2 $Old$ November 3rd, 2009, 06:36 AM

That depends.

It's a good idea to ask for specific examples rather than really general questions. It's hard to answer statements like the one you wrote here.

The answer to your question depends on if you mean by hand vs. computer, only basic algebra versus logarithms, etc.

Stuck Man · #3 $Old$ November 3rd, 2009, 08:57 AM

e^x + 3e^y = 3
e^2x - 9e^2y = 6

The answer has to be expressed as logs to base e.

I have the answers and they seem to be correct.

alexmahone · #4 $Old$ November 3rd, 2009, 09:09 AM

Quote:

Originally Posted by Stuck Man $View Post$

e^x + 3e^y = 3
e^2x - 9e^2y = 6

The answer has to be expressed as logs to base e.

I have the answers and they seem to be correct.

$e^x+3e^y=3$

$e^{2x}-9e^{2y}=6$

$(e^x)^2-(3e^y)^2=6$

$(e^x+3e^y)(e^x-3e^y)=6$

$3(e^x-3e^y)=6$

$e^x-3e^y=2$

$2e^x=5$

$e^x=\frac{5}{2}$

$x=ln (\frac{5}{2})$

$6e^y=1$

$e^y=\frac{1}{6}$

$y=ln(\frac{1}{6})=-ln 6$

Stuck Man · #5 $Old$ November 4th, 2009, 05:42 AM

Thanks.